vault backup: 2025-10-31 16:26:48

conductor-sizing.md

@@ -88,6 +88,8 @@ are in contradiction to this idea.
 
 Given a minimum ampacity, find all valid configurations.
 
+[[nfpa-70_310_conductors_for_general_wiring]]
+
 > [!cite] NEC Article 310 (emphasis added)
 > #### 310.10(H) Conductors in Parallel.
 > ##### (1) General.
@@ -162,31 +164,31 @@ $$
 > [!info] 3-Phase Voltage
 >
 > $$
-> V_{LL} = \sqrt{3} \times V_{LN}, \quad V_{LN} = \frac{V_{LL}}{\sqrt{3}}
+> V_{L} = \sqrt{3} \times V_{P}, \quad V_{P} = \frac{V_{L}}{\sqrt{3}}
 > $$
 
 3% allowable voltage drop for a 120V line-to-neutral load:
 
 $$
-\text{Max} \Delta V = 0.03 \times 120 \text{V}_{LN} = 3.6 \text{V}_{LN}
+\text{Max} \Delta V = 0.03 \times 120 \text{V}_{P} = 3.6 \text{V}_{P}
 $$
 
 3% allowable voltage drop for a 208V line-to-line load:
 
 $$
-\text{Max} \Delta V = 0.03 \times 208 \text{V}_{LL} = 6.24 \text{V}_{LL}
+\text{Max} \Delta V = 0.03 \times 208 \text{V}_{L} = 6.24 \text{V}_{L}
 $$
 
 #### Line to Neutral Loads
 
 $$
-\Delta V_{LN} = I \times Z \times 2L
+\Delta V_{P} = I \times Z \times 2L
 $$
 
 #### Line to Line Loads
 
 $$
-\Delta V_{LL} = \sqrt{3} \times \left( I \times Z \times 2L \right)
+\Delta V_{L} = \sqrt{3} \times \left( I \times Z \times 2L \right)
 $$
 
 #### 3-Phase Loads